Fourier Series 2

2010-08-04 9:14 pm
Show the following function is even:

圖片參考:http://i707.photobucket.com/albums/ww74/stevieg90/03-97.jpg

(http://i707.photobucket.com/albums/ww74/stevieg90/03-97.jpg)

Answer part (b) only, using (a):


圖片參考:http://i707.photobucket.com/albums/ww74/stevieg90/02-122.jpg

(http://i707.photobucket.com/albums/ww74/stevieg90/02-122.jpg)
更新1:

Lotus, 你還沒有回答第一題喔!

更新2:

最上邊那題: Show that the function is even...

回答 (1)

2010-08-05 8:41 am
✔ 最佳答案
f(x)每一段斜率均為1, 而g(x)每一段斜率均為 -1,
因此將f(x)改為 f(-x), 再下移π單位即可
故 g(x)=f(-x) -π
註: g(x)與 f(-x)-π 兩函數只在不連續點(x=2nπ)有差異,但Foruier series收斂
至[g(x+)+g(x-)]/2, 故g(x)與 f(-x)-π 的Fourier series相同

(i) g(x)的Fourier series為 -2Σ[n=1~∞] sin(-nx)/n= 2Σ[n=1~∞] sin(nx)/n
(ii) 令x=π/2 (註: g(x)在x=π/2處連續)
故g(π/2)=2Σ[n=1~∞] sin(nπ/2)/n
π/2=2[1/1- 1/3+1/5-1/7+...] = - 2Σ[n=1~∞] (-1)^n/(2n-1)
故 -π/4 =Σ[n=1~∞] (-1)^n/(2n-1)

2010-08-07 01:19:33 補充:
Answer part (b) only, using (a) ???

2010-08-07 22:30:36 補充:
Q1: f(x)=
{ x, 0 <= x < π
{ 2π-x, π <= x < 2π
只定義於0~2π, 故不是even fn.
若加上條件 f(x+2π)=f(x): 週期2π,則為even fn.
判別方法:作圖, 對稱於y軸, 故為even fn.


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